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Chapter 14 Chemical Equilibrium Chapter 14 Lecture

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Chapter 14 Chemical Equilibrium Chapter 14 Lecture
Chapter 14 Lecture
Lecture Presentation
Chapter 14
Chemical
Equilibrium
Sherril Soman
Grand Valley State University
© 2014 Pearson Education, Inc.
Hemoglobin
• Hemoglobin is a protein (Hb),
found in red blood cells, that reacts
with O2.
– It enhances the amount of O2
that can be carried through
the bloodstream.
Hb + O2  HbO2
– The  is used to describe
a process that is in dynamic
equilibrium.
© 2014 Pearson Education, Inc.
Hemoglobin Equilibrium System
Hb + O2  HbO2
• The concentrations of Hb, O2, and HbO2 are all
interdependent.
• The relative amounts of Hb, O2, and HbO2 at equilibrium
are related to a constant called the equilibrium
constant, K.
– A large value of K indicates a high concentration of
products at equilibrium.
• Changing the concentration of any one of these
necessitates changes the other concentrations to restore
equilibrium.
© 2014 Pearson Education, Inc.
O2 Transport
• In the lungs:
– High concentration of O2
– The equilibrium shifts to
the right
– Hb and O2 combine to
make more HbO2
© 2014 Pearson Education, Inc.
Insert cartoon at
top on page 650:
shifting of reaction
to the right
O2 Transport
• In the muscles:
– Low concentration of O2,
– The equilibrium shifts to
the right
– HbO2 breaks down
(dissociates) increasing
the amount of free O2.
© 2014 Pearson Education, Inc.
Insert cartoon at
middle on page
650 : shifting of
reaction to the left
Fetal Hemoglobin, HbF HbF + O2  HbFO2
• Fetal hemoglobin’s
equilibrium constant is
larger than adult
hemoglobin’s constant.
Hb
+
• Fetal hemoglobin is more
efficient at binding O2.
• O2 is transferred to the fetal
hemoglobin from the
mother’s hemoglobin in the
placenta.
© 2014 Pearson Education, Inc.
O2
HbO2 2
 HbO
O2
HbF +
O2
HbFO22
 HbFO
Oxygen Exchange between Mother and Fetus
© 2014 Pearson Education, Inc.
Arrow Conventions
• Chemists commonly use two kinds of
arrows in reactions to indicate the
degree of completion of the reactions.
• A single arrow indicates all the reactant
molecules are converted to product
molecules at the end.
• A double arrow indicates the reaction
stops when only some of the reactant
molecules have been converted into
products.
–  in these notes
© 2014 Pearson Education, Inc.
Reaction Dynamics
• When a reaction starts, the reactants are consumed
and products are made.
– The reactant concentrations decrease and the product
concentrations increase.
– As reactant concentration decreases, the forward reaction rate
decreases.
• Eventually, the products can react to re-form some of
the reactants, assuming the products are not allowed to
escape.
– As product concentration increases, the reverse reaction rate
increases.
• Processes that proceed in both the forward and reverse
direction are said to be reversible.
reactants  products
© 2014 Pearson Education, Inc.
Dynamic Equilibrium
• As the forward reaction slows and the reverse reaction
accelerates, eventually they reach the same rate.
• Dynamic equilibrium is the condition wherein the rates of
the forward and reverse reactions are equal.
• Once the reaction reaches equilibrium, the concentrations
of all the chemicals remain constant because the
chemicals are being consumed and made at the
same rate.
© 2014 Pearson Education, Inc.
H2(g) + I2(g)  2 HI(g)
At time 0, there are only reactants in the mixture,
so only the forward reaction can take place.
[H2] = 8, [I2] = 8, [HI] = 0
© 2014 Pearson Education, Inc.
H2(g) + I2(g)  2 HI(g)
At time 16, there are both reactants and products in the
mixture, so both the forward reaction and reverse
reaction can take place.
[H2] = 6, [I2] = 6, [HI] = 4
© 2014 Pearson Education, Inc.
H2(g) + I2(g)  2 HI(g)
At time 32, there are now more products than reactants in
the mixture, the forward reaction has slowed down as the
reactants run out, and the reverse reaction accelerated.
[H2] = 4, [I2] = 4, [HI] = 8
© 2014 Pearson Education, Inc.
H2(g) + I2(g)  2 HI(g)
At time 48, the amounts of products and reactants in the
mixture haven’t changed; the forward and reverse reactions
are proceeding at the same rate. It has reached equilibrium.
© 2014 Pearson Education, Inc.
H2(g) + I2(g)  2 HI(g)
As the concentration of product increases and the
concentrations of reactants decrease, the rate of the
forward reaction slows down, and the rate of the reverse
reaction speeds up.
© 2014 Pearson Education, Inc.
H2(g) + I2(g)  2 HI(g)
At dynamic equilibrium, the rate of the forward reaction is
equal to the rate of the reverse reaction.
The concentrations of reactants and products no longer
change.
© 2014 Pearson Education, Inc.
Equilibrium  Equal
• The rates of the forward and reverse reactions are equal
at equilibrium.
• But that does not mean the concentrations of reactants
and products are equal.
• Some reactions reach equilibrium only after almost all
the reactant molecules are consumed; we say the
position of equilibrium favors the products.
• Other reactions reach equilibrium when only a small
percentage of the reactant molecules are consumed; we
say the position of equilibrium favors the reactants.
© 2014 Pearson Education, Inc.
An Analogy: Population Changes
When Country A citizens feel overcrowded,
some will emigrate to Country B .
© 2014 Pearson Education, Inc.
An Analogy: Population Changes
However, after a time, emigration will occur in
both directions at the same rate, leading to
populations in Country A and Country B that are
constant, but not necessarily equal.
© 2014 Pearson Education, Inc.
Equilibrium Constant
• Even though the concentrations of reactants and products
are not equal at equilibrium, there is a relationship
between them.
• The relationship between the chemical equation and the
concentrations of reactants and products is called the law
of mass action.
© 2014 Pearson Education, Inc.
Equilibrium Constant
• For the general equation aA + bB  cC + dD, the law of
mass action gives the relationship below.
– The lowercase letters represent the coefficients of the
balanced chemical equation.
– Always products over reactants
• K is called the equilibrium constant.
– Unitless
© 2014 Pearson Education, Inc.
Writing Equilibrium Constant Expressions
• So, for the reaction
2 N2O5(g)  4 NO2(g) + O2(g)
the equilibrium constant expression is as follows:
© 2014 Pearson Education, Inc.
What Does the Value of Keq Imply?
• When the value of Keq >> 1, when the reaction
reaches equilibrium there will be many more
product molecules present than reactant
molecules.
• The position of equilibrium favors products.
© 2014 Pearson Education, Inc.
What Does the Value of Keq Imply?
• When the value of Keq << 1, when the reaction
reaches equilibrium there will be many more
reactant molecules present than product
molecules.
• The position of equilibrium favors reactants.
© 2014 Pearson Education, Inc.
A Large Equilibrium Constant
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A Small Equilibrium Constant
© 2014 Pearson Education, Inc.
Relationships between K and Chemical
Equations
• When the reaction is written backward, the
equilibrium constant is inverted.
For the reaction aA + bB  cC + dD
the equilibrium constant expression
is as follows:
© 2014 Pearson Education, Inc.
For the reaction cC + dD  aA + bB
the equilibrium constant expression
is as follows:
Relationships between K and Chemical
Equations
• When the coefficients of an equation are multiplied
by a factor, the equilibrium constant is raised to
that factor.
For the reaction aA + bB  cC
the equilibrium constant
expression is as follows:
© 2014 Pearson Education, Inc.
For the reaction 2aA + 2bB  2cC
the equilibrium constant
expression is as follows:
Relationships between K and Chemical
Equations
• When you add equations to get a new equation,
the equilibrium constant of the new equation is the
product of the equilibrium constants of the old
equations.
For the reactions (1) aA  bB
and (2) bB  cC the equilibrium
constant expressions are as
follows:
© 2014 Pearson Education, Inc.
For the reaction aA  cC
the equilibrium constant
expression is as follows:
Equilibrium Constants for Reactions
Involving Gases
• The concentration of a gas in a mixture is proportional to
its partial pressure.
• Therefore, the equilibrium constant can be expressed as
the ratio of the partial pressures of the gases.
• For aA(g) + bB(g)  cC(g) + dD(g) the equilibrium
constant expressions are as follows:
or
© 2014 Pearson Education, Inc.
Kc and Kp
• In calculating Kp, the partial pressures are always in atm.
• The values of Kp and Kc are not necessarily the same
because of the difference in units.
– Kp = Kc when Dn = 0
• The relationship between them is as follows:
Dn is the difference between the number of
moles of reactants and moles of products.
© 2014 Pearson Education, Inc.
Deriving the Relationship between Kp
and Kc
© 2014 Pearson Education, Inc.
Deriving the Relationship between Kp
and Kc
for aA(g) + bB(g)  cC(g) + dD(g)
substituting
© 2014 Pearson Education, Inc.
Heterogeneous Equilibria
• The concentrations of pure solids and pure liquids do not
change during the course of a reaction.
• Because their concentration doesn’t change, solids and
liquids are not included in the equilibrium constant
expression.
• For the reaction
the equilibrium constant expression is as follows:
© 2014 Pearson Education, Inc.
Heterogeneous Equilibria
The amount of C is
different, but the
amounts of CO and
CO2 remain the same.
Therefore, the amount
of C has no effect on
the position of
equilibrium.
© 2014 Pearson Education, Inc.
Calculating Equilibrium Constants from
Measured Equilibrium Concentrations
• The most direct way of finding the equilibrium constant is to
measure the amounts of reactants and products in a
mixture at equilibrium.
• The equilibrium mixture may have different amounts of
reactants and products, but the value of the equilibrium
constant will always be the same, as long as the
temperature is kept constant.
– The value of the equilibrium constant is independent of
the initial amounts of reactants and products.
© 2014 Pearson Education, Inc.
Initial and Equilibrium Concentrations for
H2(g) + I2(g)  2HI(g) at 445 °C
© 2014 Pearson Education, Inc.
Calculating Equilibrium Concentrations
• Stoichiometry can be used to determine the equilibrium
concentrations of all reactants and products if you know
initial concentrations and one equilibrium concentration.
• Use the change in the concentration of the material that
you know to determine the change in the other
chemicals in the reaction.
© 2014 Pearson Education, Inc.
The Reaction Quotient
• If a reaction mixture containing both reactants and
products is not at equilibrium, how can we determine in
which direction it will proceed?
• The answer is to compare the current concentration
ratios to the equilibrium constant.
• The concentration ratio of the products (raised to the
power of their coefficients) to the reactants (raised to the
power of their coefficients) is called the reaction
quotient, Q.
© 2014 Pearson Education, Inc.
The Reaction Quotient
For the gas phase reaction
aA + bB  cC + dD
the reaction quotient is as follows:
© 2014 Pearson Education, Inc.
The Reaction Quotient: Predicting the
Direction of Change
• If Q > K, the reaction will proceed fastest in the reverse
direction.
– The products will decrease and reactants will increase.
© 2014 Pearson Education, Inc.
The Reaction Quotient: Predicting the
Direction of Change
• If Q < K, the reaction will proceed fastest in the forward
direction.
– The products will increase and reactants will decrease.
© 2014 Pearson Education, Inc.
The Reaction Quotient: Predicting the
Direction of Change
• If Q = K, the reaction is at equilibrium
– The products and reactants will not change.
• If a reaction mixture contains just reactants, then Q = 0, and
the reaction will proceed in the forward direction.
• If a reaction mixture contains just products, then Q = ∞, and
the reaction will proceed in the reverse direction.
© 2014 Pearson Education, Inc.
Finding Equilibrium Concentrations
When Given the Equilibrium Constant
and Initial Concentrations or Pressures
Step 1: Decide in which direction the reaction will proceed.
– Compare Q to K.
Step 2: Define the changes of all materials in terms of x.
– Use the coefficient from the chemical equation as the
coefficient of x.
– The x change is + for materials on the side the reaction is
proceeding toward.
– The x change is  for materials on the side the reaction is
proceeding away from.
Step 3: Solve for x.
– For second order equations, take square roots of both sides or use
the quadratic formula.
– Simplify and approximate answer for very large or small equilibrium
constants, if possible.
© 2014 Pearson Education, Inc.
Approximations to Simplify the Math
• When the equilibrium constant is very small, the
position of equilibrium favors the reactants.
• For relatively large initial concentrations of
reactants, the reactant concentration will not change
significantly when it reaches equilibrium.
– assuming the reaction is proceeding forward
– The [X]equilibrium = ([X]initial  ax)  [X]initial
• We are approximating the equilibrium concentration of
reactant to be the same as the initial concentration.
© 2014 Pearson Education, Inc.
Checking the Approximation and Refining
as Necessary
• We can check our approximation by
comparing the approximate value of x to the
initial concentration.
• If the approximate value of x is less than 5%
of the initial concentration, the approximation
is valid.
© 2014 Pearson Education, Inc.
Disturbing and Restoring Equilibrium
• Once a reaction is at equilibrium, the
concentrations of all the reactants and products
remain the same.
• However, if the conditions are changed, the
concentrations of all the chemicals will change
until equilibrium is restored.
• The new concentrations will be different, but the
equilibrium constant will be the same, unless you
change the temperature.
© 2014 Pearson Education, Inc.
Le Châtelier’s Principle
• Le Châtelier's principle guides us in predicting the effect
various changes in conditions have on the position of
equilibrium.
• It says that if a system at equilibrium is disturbed, the
position of equilibrium will shift to minimize the disturbance.
– Disturbances all involve making the system open.
© 2014 Pearson Education, Inc.
An Analogy: Population Changes
When the populations of Country A and Country B
are in equilibrium, the emigration rates between the
two countries are equal so the populations stay constant.
© 2014 Pearson Education, Inc.
An Analogy: Population Changes
When an influx of population enters Country B
from somewhere outside Country A, it disturbs the
equilibrium established between Country A and Country B.
© 2014 Pearson Education, Inc.
An Analogy: Population Changes
The result will be people moving from Country B into
Country A faster than people moving from Country A into
Country B.
This will continue until a new equilibrium between the
populations is established; the new populations will have
different numbers of people than the old ones.
© 2014 Pearson Education, Inc.
Disturbing Equilibrium: Adding or
Removing Reactants
• After equilibrium is established, a reactant is added,
as long as the added reactant is included in the
equilibrium constant expression.
• That is, not a solid or liquid
• How will this affect the rate of the forward reaction?
• How will it affect the rate of the reverse reaction?
• How will it affect the value of K?
© 2014 Pearson Education, Inc.
Disturbing Equilibrium: Adding Reactants
• Adding a reactant initially increases the rate of the
forward reaction, but has no initial effect on the rate
of the reverse reaction.
• The reaction proceeds to the right until equilibrium
is restored.
• At the new equilibrium position, you will have more
of the products than before, less of the non-added
reactants than before, and less of the added
reactant.
– But you will not have as little of the added reactant as
you had before the addition.
• At the new equilibrium position, the concentrations
of reactants and products will be such that the value
of the equilibrium constant is the same.
© 2014 Pearson Education, Inc.
Disturbing Equilibrium: Adding or
Removing Reactants
• After equilibrium is established, a reactant is
removed, as long as the added reactant is included in
the equilibrium constant expression.
• That is, not a solid or liquid
• How will this affect the rate of the forward reaction?
• How will it affect the rate of the reverse reaction?
• How will it affect the value of K?
© 2014 Pearson Education, Inc.
Disturbing Equilibrium: Removing Reactants
• Removing a reactant initially decreases the rate of
the forward reaction, but has no initial effect on the
rate of the reverse reaction.
– So the reaction is going faster in reverse.
• The reaction proceeds to the left until equilibrium is
restored.
• At the new equilibrium position, you will have less
of the products than before, more of the nonremoved reactants than before, and more of the
removed reactant.
– But you will not have as much of the removed reactant
as you had before the removal.
• At the new equilibrium position, the concentrations
of reactants and products will be such that the
value of the equilibrium constant is the same.
© 2014 Pearson Education, Inc.
The Effect of Concentration Changes on
Equilibrium
• Adding a reactant will decrease the amounts of the
other reactants and increase the amount of the
products until a new position of equilibrium is found
that has the same K.
• Removing a product will increase the amounts of the
other products and decrease the amounts of the
reactants.
– You can use this to drive a reaction to completion!
• Equilibrium shifts away from the side with added
chemicals or toward the side with removed
chemicals.
– Remember, adding more of a solid or liquid does not
change its concentration; therefore, it has no effect on the
equilibrium.
© 2014 Pearson Education, Inc.
The Effect of Concentration Changes on
Equilibrium
When NO2 is added, some of it
combines to make more N2O4.
© 2014 Pearson Education, Inc.
The Effect of Concentration Changes
on Equilibrium
When N2O4 is added, some of it
decomposes to make more NO2.
© 2014 Pearson Education, Inc.
The Effect of Concentration Changes
on Equilibrium
When N2O4 is added, some of it
decomposes to make more NO2.
© 2014 Pearson Education, Inc.
The Effect of Adding a Gas to a
Gas Phase Reaction at Equilibrium
• Adding a gaseous reactant increases its
partial pressure, causing the equilibrium to
shift to the right.
– Increasing its partial pressure increases its
concentration.
– It does not increase the partial pressure of the
other gases in the mixture.
• Adding an inert gas to the mixture has no
effect on the position of equilibrium.
– It does not affect the partial pressures of the
gases in the reaction.
© 2014 Pearson Education, Inc.
Effect of Volume Change on Equilibrium
• Decreasing the volume of the container increases the
concentration of all the gases in the container.
– It increases their partial pressures.
– It does not change the concentrations of solutions!
• If their partial pressures increase, then the total
pressure in the container will increase.
• According to Le Châtelier’s Principle, the equilibrium
should shift to remove that pressure.
• The way the system reduces the pressure is to reduce
the number of gas molecules in the container.
• When the volume decreases, the equilibrium shifts
to the side with fewer gas molecules.
© 2014 Pearson Education, Inc.
Disturbing Equilibrium: Changing the
Volume
• After equilibrium is established, the container volume is
decreased.
• How will it affect the concentration of solids, liquid,
solutions, and gases?
• How will this affect the total pressure of solids, liquid,
and gases?
• How will it affect the value of K?
© 2014 Pearson Education, Inc.
Disturbing Equilibrium: Reducing the
Volume
• Decreasing the container volume will increase the total
pressure.
– Boyle’s law
– If the total pressure increases, the partial pressures of all the
gases will increase—Dalton’s law of partial pressures.
• Because the total pressure increases, the position of
equilibrium will shift to decrease the pressure by removing
gas molecules.
– Shift toward the side with fewer gas molecules
• At the new equilibrium position, the partial pressures of
gaseous reactants and products will be such that the value
of the equilibrium constant is the same.
© 2014 Pearson Education, Inc.
The Effect of Volume Changes
on Equilibrium
Left side of
figure
14.11
© 2014 Pearson Education, Inc.
Because there are more
gas molecules on the
reactants side of the
reaction, when the pressure
is increased, the position of
equilibrium shifts toward the
side with fewer molecules
to decrease the pressure.
The Effect of Volume Changes
on Equilibrium
Right side
When the pressure is
decreased by increasing the of figure
14.11
volume, the position of
equilibrium shifts toward the
side with the greater
number of molecules—the
reactant side.
© 2014 Pearson Education, Inc.
The Effect of Temperature Changes on
Equilibrium Position
• Exothermic reactions release energy and
endothermic reactions absorb energy
• Writing heat as a product in an exothermic
reaction or as a reactant in an endothermic
reaction, helps us use Le Châtelier’s principle to
predict the effect of temperature changes, even
though heat is not matter and not written in a
proper equation.
© 2014 Pearson Education, Inc.
The Effect of Temperature Changes on
Equilibrium for Exothermic Reactions
• For an exothermic reaction, heat is a product.
• Increasing the temperature is like adding heat.
• According to Le Châtelier’s principle, the
equilibrium will shift away from the added heat.
© 2014 Pearson Education, Inc.
The Effect of Temperature Changes on
Equilibrium for Exothermic Reactions
• Adding heat to an exothermic reaction will
decrease the concentrations of products and
increase the concentrations of reactants.
• Adding heat to an exothermic reaction will
decrease the value of K.
• How will decreasing the temperature affect the
system?
© 2014 Pearson Education, Inc.
The Effect of Temperature Changes on
Equilibrium for Endothermic Reactions
• For an endothermic reaction, heat is a reactant
• Increasing the temperature is like adding heat
• According to Le Châtelier’s Principle, the
equilibrium will shift away from the added heat
© 2014 Pearson Education, Inc.
The Effect of Temperature Changes on
Equilibrium for Endothermic Reactions
• Adding heat to an endothermic reaction will
decrease the concentrations of reactants and
increase the concentrations of products.
• Adding heat to an endothermic reaction will
increase the value of K.
• How will decreasing the temperature affect the
system?
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The Effect of Temperature Changes on
Equilibrium
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Not Changing the Position of Equilibrium:
The Effect of Catalysts
• Catalysts provide an alternative, more
efficient mechanism.
• Catalysts work for both forward and
reverse reactions.
• Catalysts affect the rate of the forward and
reverse reactions by the same factor.
• Therefore, catalysts do not affect the
position of equilibrium.
© 2014 Pearson Education, Inc.
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